Question

\[\int\frac{x + 1}{x^2 + 4x + 5} \text{  dx}\]

Answer 1

\[\text{ Let  I } = \int\frac{\left( x + 1 \right)}{x^2 + 4x + 5}dx\]

\[\text{ and   let} \left( x + 1 \right) = A\frac{d}{dx}\left( x^2 + 4x + 5 \right) + B\]

\[ \Rightarrow x + 1 = A \left( 2x + 4 \right) + B\]

\[ \Rightarrow x + 1 = \left( 2A \right)x + 4A + B\]

\[\text{Equating the coefficients of like terms}\]

\[2A = 1\]

\[ \Rightarrow A = \frac{1}{2}\]

\[\text{ and }\ 4A + B = 1\]

\[ \Rightarrow 4 \times \frac{1}{2} + B = 1\]

\[ \Rightarrow B = - 1\]

\[ \therefore \left( x + 1 \right) = \frac{1}{2} \left( 2x + 4 \right) - 1\]

\[ \therefore I = \int\left[ \frac{\frac{1}{2}\left( 2x + 4 \right) - 1}{x^2 + 4x + 5} \right]dx\]

\[ = \frac{1}{2}\int\frac{\left( 2x + 4 \right)}{x^2 + 4x + 5}dx - \int\frac{1}{x^2 + 4x + 5}dx\]

\[\text{ Putting  x}^2 + 4x + 5 = t\]

\[ \Rightarrow \left( 2x + 4 \right) dx = dt\]

\[ \therefore I = \frac{1}{2}\int\frac{1}{t}dt - \int\frac{1}{x^2 + 4x + 4 + 1}dx\]

\[ = \frac{1}{2}\int\frac{dt}{t} - \int\frac{1}{\left( x + 2 \right)^2 + 1^2}dx \]

\[ = \frac{1}{2} \text{ ln } \left| t \right| - \tan^{- 1} \left( \frac{x + 2}{1} \right) + C............. \left[ \because \int\frac{1}{x^2 + a^2}dx = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right]\]

\[ = \frac{1}{2} \text{ ln }\left| x^2 + 4x + 5 \right| - \tan^{- 1} \left( x + 2 \right) + C ...................\left[ \because t = x^2 + 4x + 5 \right]\]